Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Can non-constant polynomials over algebraically closed fields never vanish? No, right?

Thoughts: Take $f(x,y)$, suppose that for whatever $y'$ we plug in for $y$ we find that $f(x,y')=c$, that is it kills the $x$ term. This would mean that $f(x,y)$ is really just a polynomial in $y$ and boom we are done, because a polynomial in one variable has a root in a a.c. field. Suppose that's not the case, i.e. if we plug in some $y'$ we just get a polynomial in $x$, then by same reasoning, the a.c. guarantees a root. Same argument applies to more variables.

So the main task is: given two affine varieties that are disjoint, $V$ and $V'$, construct a regular function that vanishes on $V$ and is constant $c$ on $V'$. In the simplest case we just have that each of these varieties is defined by a single polynomial each, $f$ and $g$, that differ by a constant, such that $f-g=d$. Then the polynomial we want would be $cf/(f-g)$, because for a point on $V$ we know $f=0$ so this gives $0/(-g)$ (and we know $g\neq0$ on $V$ since $V\cap V'= \emptyset$) and for a point on $V'$ we know $g=0$ which gives $cf/f$ (and we know $f\neq0$ on $V'$ since $V\cap V'= \emptyset$) so this is just $c$. However, I am not sure how to generalize this to varieties defined by polynomials that aren't just different by additive constants and then generalize it more so to varieties defined by multiple polynomials.

I mean, can we deduce from the fact if $f-g$ is a polynomial (rather than a constant) that the varieties $f$ and $g$ define intersect? If $f-g$ is a polynomial, then it must vanish thus there are points where $f=g$. However, these points don't necessarily need to be points where $f=g=0$, and thus I don't think the varieties need to intersect necessarily, but I'm not sure.


share|cite|improve this question
Could you please state precisely what your question is? If it is "do affine subvarieties of $\mathbb A^n_k$ necessarily intersect", the answer is "no": parallel lines have been with us for a few thousand years. – Georges Elencwajg Mar 31 '13 at 6:07
The answer to the first question is no by induction on the number of variables. I don't understand what the rest of the question has to do with the first question. – Qiaochu Yuan Mar 31 '13 at 6:26
It is easy to show by induction on the number of a variables that a polynomial function over an infinite field does not vanish identically (proven on this site many times). An algebraically closed field is necessarily infinite, so... – Jyrki Lahtonen Mar 31 '13 at 7:15
up vote 4 down vote accepted

I'm answering your question in your second paragraph.

Suppose $k$ is an algebraically closed field. If $f,g\in k[x,y]$ are nonzero polynomials without a common zero point, then $V(f,g)$ is empty. By Hilbert's Nullstellensatz, $\sqrt{(f,g)}=(1)$, so $(f,g)=(1)$, which means that there exists $u,v\in k[x,y]$ such that $uf-vg=1$. For any constant $c$ take $h:=cuf=cvg+c$, then $h$ vanishes $V:=V(f)$ and is constant $c$ on $V':=V(g)$.

The idea is that if the difference of $f$ and $g$ is a non-constant polynomial, we find $u$ and $v$ such that the difference of $uf$ and $vg$ is 1, while $uf$ also vanishes on $V$ and $vg$ also vanishes on $V'$.

I think this result is also true for any two disjoint affine varieties $V:=V(I),V':=V(I')$ in $\mathbb{A}_k^n$, since you can modify the proof above substituting $f,g$ by $I,I'$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.